# How do you sketch the graph y=x^4-2x^3+2x using the first and second derivatives?

Jun 12, 2018

See process below

#### Explanation:

1.- Domain is $\mathbb{R}$
2.- Analyze the roots of equation ${x}^{4} - 2 {x}^{3} + 2 x = 0$ in order to find axis intercepts

Notice that ${x}^{4} - 2 {x}^{3} + 2 x = x \left({x}^{3} - 2 {x}^{2} + 2\right) = 0$ thus $\left(0 , 0\right)$ is a passing point. There is no more integer roots

3.- First derivative f´(x)=4x^3-6x^2+2=(x-1)^2(4x+2)

Analyze the sign of derivative. ${\left(x - 1\right)}^{2}$ is allways positive, then the sign of derivative dependes only of sign of $4 x + 2$

$4 x + 2 \ge 0$
$x \ge - \frac{1}{2}$

Then function is increasing in $\left(- \frac{1}{2} , \infty\right)$ and decreasing in $\left(- \infty , - \frac{1}{2}\right)$ then in $x = - \frac{1}{2}$ has a minimum

Derivative is zero in $x = - \frac{1}{2}$ and $x = 1$ but $x = 1$ is not a maximum neither a minimum because

f´´(x)=12x^2-12x=12x(x-1). then f´´ become $0$ only in $x = 0$ and $x = 1$. Thus inflection points.

A sketch is presented below 